Math 450 Midterm 2
1. Suppose f (0) = 3, f (1) = 1 and f (3) = 9. Using a divided dierence table, nd the
Newton form of the degree 2 polynomial P that interpolates f at these points.
Solution.
The divided dierence table is
0 3
−2
1 1
2
4
3 9
so P (x) = 3 − 2(x − 0) + 2(x − 0)(x − 1) is the Newton form of P .
1
2. Let P (x) = 2 + 3(x − 1) + 4(x − 1)(x − 2) + 5(x − 1)(x − 2)(x − 3). Using either the
Newton form given or the nested form of P , write a pseudocode to compute P (4) with a
single for loop.
Solution.
A loop for the Newton form is
y = 2, r = 1
for i = 1 : 3
r = r(4 − i)
y = y + (i + 2)r
end
A loop for the nested form is
y=5
for i = 3 : 1
y = y(4 − i) + (i + 1)
end
2
3. Let f (x) = ex be interpolated1 at points x0 , x1 , x2 , x3 , x4 in [−1, 1].
i) When the xi 's are equally spaced, give an upper bound for |f (x) − P (x)|.
e
= 640
: notice |f (5) (ξ)| ≤ e and recall
An upper bound is |f (x) − P (x)| ≤ e(1/2)
4·5
Q4
5
i=0 |x − xi | ≤ (1/2) 4! where 1/2 is the spacing between nodes.
5
Solution.
ii) Do the same as in part i) when the xi 's are Chebyshev nodes.
e
An upper bound is |f (x) − P (x)| ≤ e25! = 1920
: notice |f (5) (ξ)| ≤ e and recall
−4
when xi 's are Chebyshev nodes in [−1, 1].
i=0 |x − xi | ≤ 2
−4
Solution.
Q4
1 Recall:
when
f
is interpolated by a degree
|f (x) − P (x)| =
≤n
polynomial
n
|f (n+1) (ξ)| Y
|x − xi |,
(n + 1)! i=0
3
P
at distinct points
x 0 ≤ x ≤ xn .
x0 , x1 , . . . , xn ,
4. Show the approximation f 0 (x) ≈
Solution.
1
[f (x
4h
+ 3h) − f (x − h)] has error O(h).
Write
1
f (x + 3h) = f (x) + f 0 (x)(3h) + f 00 (x)(3h)2 + . . .
2
1
f (x − h) = f (x) + f 0 (x)(−h) + f 00 (x)(−h)2 + . . . ,
2
subtract and divide by 4h to get
f (x + 3h) − f (x − h)
= f 0 (x) + f 00 (x)h + . . . = f 0 (x) + f 00 (ξ)h.
4h
4
5. Let P (x) = c0 + c1 x + c2 x2 . Write a linear system corresponding to the equations
P (0) = 1, P (1) = 2, P (−1) = 2 and P (2) = 0, and explain why there is no solution for
the unknown ci 's.
Solution.
The linear system is


 
1 0 0  
1
 1 1 1  c0
2

   
1 −1 1 c1 = 2 .
c2
1 2 4
0
The rst equation shows c0 = 1, so the second and third equations show that c1 +c2 = 1 =
c2 − c1 , so that c1 = 0 and c2 = 1. But then the fourth equation is 0 = c0 + 2c1 + 4c2 = 5,
which is false. The linear system is overdetermined: there are 4 independent equations,
but only 3 unknowns.
5